两个整数的二进制表示中不匹配的位数
给定两个整数(小于 2^31) A 和 b，任务是找出二进制表示中不同的位数。
例:
Input : A = 12, B = 15
Output : Number of different bits : 2
Explanation: The binary representation of
12 is 1100 and 15 is 1111.
So, the number of different bits are 2.
Input : A = 3, B = 16
Output : Number of different bits : 3
方法:
- Loop from' 0' to' 31', shift the bits of A and B to the right by' I' bit, and then check whether the bits of' 0' bit are different.
- If the bits are different, the count is increased.
- Since the number is less than 2.31, we only need to cycle' 32' times, that is, from' 0' to' 31'.
- If we "AND" 1 the number bit by bit, we can get the first digit.
- Display the count at the end of the loop.
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// compute number of different bits
void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
cout << "Number of different bits : " << count << endl;
}
// Driver code
int main()
{
int A = 12, B = 15;
// find number of different bits
solve(A, B);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG {
// compute number of different bits
static void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
System.out.println("Number of different bits : " + count);
}
// Driver code
public static void main (String[] args) {
int A = 12, B = 15;
// find number of different bits
solve(A, B);
}
}
// this code is contributed by anuj_67..
Python 3
# Python3 implementation of the approach
# compute number of different bits
def solve( A, B):
count = 0
# since, the numbers are less than 2^31
# run the loop from '0' to '31' only
for i in range(0,32):
# right shift both the numbers by 'i' and
# check if the bit at the 0th position is different
if ((( A >> i) & 1) != (( B >> i) & 1)):
count=count+1
print("Number of different bits :",count)
# Driver code
A = 12
B = 15
# find number of different bits
solve( A, B)
# This code is contributed by ihritik
C
// C# implementation of the approach
using System;
class GFG
{
// compute number of different bits
static void solve(int A, int B)
{
int count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for (int i = 0; i < 32; i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if (((A >> i) & 1) != ((B >> i) & 1)) {
count++;
}
}
Console.WriteLine("Number of different bits : " + count);
}
// Driver code
public static void Main()
{
int A = 12, B = 15;
// find number of different bits
solve(A, B);
}
}
// This code is contributed by ihritik
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// compute number of different bits
function solve($A, $B)
{
$count = 0;
// since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for ($i = 0; $i < 32; $i++) {
// right shift both the numbers by 'i' and
// check if the bit at the 0th position is different
if ((($A >> $i) & 1) != (($B >> $i) & 1)) {
$count++;
}
}
echo "Number of different bits : $count";
}
// Driver code
$A = 12;
$B = 15;
// find number of different bits
solve($A, $B);
// This code is contributed by ihritik
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Compute number of different bits
function solve(A, B)
{
var count = 0;
// Since, the numbers are less than 2^31
// run the loop from '0' to '31' only
for(i = 0; i < 32; i++)
{
// Right shift both the numbers by 'i'
// and check if the bit at the 0th
// position is different
if (((A >> i) & 1) != ((B >> i) & 1))
{
count++;
}
}
document.write("Number of different bits : " +
count);
}
// Driver code
var A = 12, B = 15;
// Find number of different bits
solve(A, B);
// This code is contributed by Rajput-Ji
</script>
Output
Number of different bits : 2
一种不同的方法使用 xor(^):
- Seek two numbers of difference or (), such as A and B. 。
- And let A&B of their result xor () become c;
- Calculate the set number of bits in the binary representation of C (1);
- Return count;
例:
- Let A = 10 (01010) and B = 20(10100)
- XOR = 11110 is obtained after XOR of A and B .. (Check the XOR table if necessary).
- Calculating the number of 1 in XOR operation gives the count of bit difference between A and B (using Brian Knigan's algorithm).
下面是上述方法的实现:
c++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// compute number of different bits
int solve(int A, int B)
{
int XOR = A ^ B;
// Check for 1's in the binary form using
// Brian Kerninghan's Algorithm
int count = 0;
while (XOR) {
XOR = XOR & (XOR - 1);
count++;
}
return count;
}
// Driver code
int main()
{
int A = 12, B = 15;
// 12 = 1100 & 15 = 1111
int result = solve(A, B);
cout << "Number of different bits : " << result;
return 0;
}
// the code is by Samarpan Chakraborty
Java
/*package whatever //do not write package name here */
// Java implementation of the approach
import java.io.*;
class GFG {
// compute number of different bits
static int solve(int A, int B)
{
int XOR = A ^ B;
// Check for 1's in the binary form using
// Brian Kerninghan's Algorithm
int count = 0;
while (XOR > 0) {
XOR = XOR & (XOR - 1);
count++;
}
// return the count of different bits
return count;
}
public static void main(String[] args)
{
int A = 12, B = 15;
// find number of different bits
int result = solve(A, B);
System.out.println("Number of different bits : "
+ result);
}
}
// the code is by Samarpan Chakraborty
python 3
# code
def solve(A, B):
XOR = A ^ B
count = 0
# Check for 1's in the binary form using
# Brian Kernighan's Algorithm
while (XOR):
XOR = XOR & (XOR - 1)
count += 1
return count
result = solve(3, 16)
# 3 = 00011 & 16 = 10000
print("Number of different bits : ", result)
# the code is by Samarpan Chakraborty
c
// C# program for the above approach
using System;
class GFG {
// compute number of different bits
static int solve(int A, int B)
{
int XOR = A ^ B;
// Check for 1's in the binary form using
// Brian Kerninghan's Algorithm
int count = 0;
while (XOR > 0) {
XOR = XOR & (XOR - 1);
count++;
}
// return the count of different bits
return count;
}
// Driver code
public static void Main()
{
int A = 12, B = 15;
// find number of different bits
int result = solve(A, B);
Console.WriteLine("Number of different bits : "
+ result);
}
}
// This code is contributed by target_2.
Javascript
<script>
/*package whatever //do not write package name here */
// JavaScript implementation of the approach
// compute number of different bits
function solve(A, B)
{
var XOR = A ^ B;
// Check for 1's in the binary form using
// Brian Kerninghan's Algorithm
var count = 0;
while (XOR > 0) {
XOR = XOR & (XOR - 1);
count++;
}
// return the count of different bits
return count;
}
var A = 12, B = 15;
// find number of different bits
var result = solve(A, B);
document.write("Number of different bits : "
+ result);
// This code is contributed by shivanisinghss2110
</script>
输出
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